# What is t such as the definite integral of 4x^3+16 is 31 if the limits of integration are t and t+1 ?

### 2 Answers | Add Yours

The indefinite integral of 4x^3 + 16 is x^4 + 16x + C

Between the limits x = t and x = t + 1, the definite integral is

=> (t + 1)^4 + 16(t + 1) + C - t^4 - 16t - C

This is equal to 31

=> (t + 1)^4 + 16(t + 1) - t^4 - 16t = 31

=> (t + 1)^4 - t^4 = 15

=> t^4 + 4t^3 + 6t^2 + 4t + 1 - t^4 = 15

=> 4t^3 + 6t^2 + 4t + 1 = 15

=> 4t^3 + 6t^2 + 4t - 14 = 0

=> (t - 1)(4t^2 + 10t + 14) = 0

=> (t - 1)(2t^2 + 5t +7) = 0

t -1 = 0

=> t1 = 1

2t^2 + 5t + 7 = 0

=> t2 = -5/4 + sqrt (25 - 56) / 4

=> t2 = -5/4 + i*(sqrt 31)/4

t3 = -5/4 - i*(sqrt 31)/4

**The values that can take are (1 , -5/4 + i*(sqrt 31)/4, -5/4 - i*(sqrt 31)/4**

Let f(x)= 4x^3 + 16

Let F(x) = Int f(x)

Then we are given that F(t+1) - F(t) = 31

Let us integrate f(x).

==> F(x)= Int 4x^3 +16 dx

==> F(x)= x^4 + 16x

==> F(t+1)= (t+1)^4 + 14(t+1)

==> F(t) = t^4 + 16t

==> F(t+1) - F(t) = 31

==> (t+1)^4 + 16(t+1) - t^4 -16t = 31

==> (t+1)^4 + 16t + 16 - t^4 - 16t = 31

Reduce similar.

==> t^4 + 4t^3+ 6t^2 +4t + 1 + 16t +16- t^4 - 16t = 31

==> 4t^3 + 6t^2 + 4t +1 + 16 = 31

==> 4t^3 + 6t^2 + 4t - 14 = 0

==> (t-1) ( 4t^2+10t+14) = 0

==> 2(t-1)(2t^2 +5t +7)= 0

==> t1= 1

==> t2= ( -5 + sqrt(25-4*2*7) /2*2

= (-5 + sqrt-31 ) / 4

= (-5/4) + sqrt31)/4 * i

==> t2= (-5/4) - sqrt31/4 *i

Then we have 3 values for t.

**==> t= { 1, (-5/4)+sqrt31/4 *i ) , (-5/4 - sqrt31/4 *i) }**